Roy Lewallen wrote:
Roy Lewallen wrote:
chuck wrote:
. . .
2. What would the ohmic losses be over a one square foot by 33 foot
path through salt water?

Let's see, salt water conductivity is about 5 S/m, which is 1.524
S/ft. So the *DC* resistance of that piece of sea water would be 1.524
* 33 / (1 * 1) ~ 50 ohms. . .

Oops. The DC resistance would be 33 / (1 * 1) / 1.524 ~ 22 ohms.

Roy Lewallen, W7EL

Thank you for the detailed response, Roy.

A couple of issues still trouble me however.

If the skin depth at 14 MHz is about 2.4 inches, can we roughly assume
that the RF resistance of that path is no less than 52.8 ohms (2.4*22
ohms)? This assumes most of the RF current would occur in the top one
inch (attenuation at one inch would be about 15 dB), and that the
resistance at 14 MHz is equal to the DC resistance.

A path one inch deep by 16 feet long (1/4 wavelength at 14 MHz) would
then have no less than 26.4 ohms resistance at 14 MHz.

Now imagine a system of multiple one foot wide by 16 feet long copper
radials on the ground with 26.4 ohm resistance distributed uniformly in
each radial. Obviously such a system will be lossy, with an average
radial resistance of 13.2 ohms.

While the analogy is a stretch, it illustrates the difficulty I am
having in understanding how seawater can be considered more efficient
than even a single slightly elevated radial, which is reported to be
less than 1 dB worse than 120 quarter wavelength buried radials
(ignoring slight pattern distortion). So even if seawater does
constitute a less lossy ground plane than a single radial (yeah, apples
and oranges, but we can weigh their juices I think) it would be better
by less than 1 dB. .

Then there is the issue of the one foot long "grounding rod" immersed in
the sea. If the above back-of-the-envelope analysis is valid, it would
seem that a even one inch long rod would be more than sufficient. If we
were dealing with a pool of liquid mercury or silver, this would have
considerable intuitive appeal for me. But the seawater model is
troubling. I imagine seawater to be a lot like earth, except more
homogeneous and with orders of magnitude higher conductivity. And I
imagine a perfect ground plane to have conductivity orders of magnitude
higher than seawater. I imagine even a modest system of copper radials
to appear more like liquid mercury than seawater does.

Where am I going astray?

73,

Chuck
NT3G